Maths Questions

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Last updated 10:00 AM on 8/26/26
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21 Terms

1
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Find the minimum value of 𝑥^2 − 4𝑥 − 12.

-16

<p>-16</p>
2
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i) (53/4) -5(x-(3/2))²

ii) x=3/2

( 3/2 , 53/4 )

<p>i) (53/4) -5(x-(3/2))²</p><p>ii) x=3/2</p><p>( 3/2 , 53/4 ) </p>
3
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bruh do NOT let the 5 marks and big words scare u its so easy

<p>bruh do NOT let the 5 marks and big words scare u its so easy</p>
4
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<p>ONLYYYY do iii (i only have the answer for that)</p>

ONLYYYY do iii (i only have the answer for that)

i didn’t really understand what bisect mean so it was confusing but now it makes sense:

When you bisect a line segment, you cut it into two equal lengths, so you want to find the midpoint

<p>i didn’t really understand what bisect mean so it was confusing but now it makes sense:</p><p>When you bisect a line segment, you cut it into two equal lengths, so you want to find the midpoint</p>
5
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6
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super easy just kinda forgot, it’s a disguised quadratic, use your own method not the one that is here, but just look at the answer


<p>super easy just kinda forgot, it’s a disguised quadratic, use your own method not the one that is here, but just look at the answer</p><p></p>
7
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8
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Remember, this isn’t a two‑number answer — it’s a range of values for k.

The discriminant becomes a quadratic expression in k, and a quadratic is positive only between its two roots.

So the original quadratic has real roots only when k lies between those two values.


but if you got it right.. well done! ;p

<p>Remember, this isn’t a two‑number answer — it’s a range of values for k. </p><p>The discriminant becomes a quadratic expression in k, and a quadratic is positive only between its two roots. </p><p>So the original quadratic has real roots only when k lies between those two values.</p><p></p><p>but if you got it right.. well done! ;p</p>
9
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10
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What does it mean if e.g. f(x) = -6 has no real solution

It doesnt intersect the curve

11
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bruh easy question too how did i get it wrong, this why you dont rush!

<p>bruh easy question too how did i get it wrong, this why you dont rush!</p>
12
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What i did wrong was i made it so y=x/2 , but its actually y=(x/2) + 1 , remember 2/2 is 1 , it doesnt cancel out into 0

<p>What i did wrong was i made it so y=x/2 , but its actually y=(x/2) + 1 , remember 2/2 is 1 , it doesnt cancel out into 0</p>
13
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14
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I was on this question for a while, I knewww that I knew how to do it but I just didn’t know where I was going wrong, turns out I factorised x²+k²x² like (2k²)x²not like (1 + k²)x² how I was meant to so yh

<p>I was on this question for a while, I knewww that I knew how to do it but I just didn’t know where I was going wrong, turns out I factorised x²+k²x² like (2k²)x²not like (1 + k²)x² how I was meant to so yh</p>
15
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<p>just do b </p><p>the centre is (4,-5) and radius is <strong>√</strong>40</p>

just do b

the centre is (4,-5) and radius is 40

How do we know which is correct?

We use the condition: “the gradient of AQ is positive.”

<p>How do we know which is correct? </p><p><span>We use the condition: </span>“the gradient of AQ is positive.”</p>
16
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ez 6 marks

<p>ez 6 marks </p>
17
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18
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btw we letting u= -4-k

<p>btw we letting u= -4-k</p>
19
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20
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21
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